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614 Find the complex frequencies that are associated with
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a b c d e 5e 4t cos 2 t sin( t + 2 ) 4e 2t sin(3t 50 ) e 3t (2 + cos 4t)
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615 Find s and V(s) if v(t) is given by
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a 5e 2t b 5e 2t cos(4t + 10 ) c 4 cos(2t 20 )
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616 Find v(t) if
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a s = 2, V = 2 0 b s = j 2, V = 12 30 c s = 4 + j 3, V = 6 10
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All the concepts and rules used in AC network analysis (see 4), such as impedance, admittance, KVL, KCL, and Th venin s and Norton s theorems, e carry over to the damped sinusoid case exactly In the complex frequency domain, the current I(s) and voltage V(s) are related by the expression V(s) = Z(s)I(s) (645)
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where Z(s) is the familiar impedance, with s replacing j We may obtain Z(s) from Z(j ) by simply replacing j by s For a resistance, R, the impedance is ZR (s) = R For an inductance, L, the impedance is ZL (s) = sL For a capacitance, C, it is ZC (s) = 1 sC (648) (647) (646)
Impedances in series or parallel are combined in exactly the same way as in the AC steady-state case, since we only replace j by s
EXAMPLE 67 Complex Frequency Notation
Problem
Use complex impedance ideas to determine the response of a series RL circuit to a damped exponential voltage
Solution
Known Quantities: Source voltage, resistor, inductor values Find: The time-domain expression for the series current, iL (t)
L = 2 H
Schematics, Diagrams, Circuits, and Given Data: vs (t) = 10e 2t cos(5t) V; R = 4 Assumptions: None Analysis: The input voltage phasor can be represented by the expression
V(s) = 10 0 V The impedance seen by the voltage source is Z(s) = R + sL = 4 + 2s Thus, the series current is: I(s) = 10 10 10 V(s) = = = = j 1 = 1 Z(s) 4 + 2s 4 + 2( 2 + j 5) j 10 2
Finally, the time-domain expression for the current is: iL (t) = e 2t cos(5t /2) A
Comments: The phasor analysis method illustrated here is completely analogous to the
method introduced in 4, with the complex frequency j (steady-state sinusoidal frequency) related by s (damped sinusoidal frequency)
Part I
Circuits
Just as frequency response functions H (j ) were de ned in this chapter, it is possible to de ne a transfer function, H (s) This can be a ratio of a voltage to a current, a ratio of a voltage to a voltage, a ratio of a current to a current, or a ratio of a current to a voltage The transfer function H (s) is a function of network elements and their interconnections Using the transfer function and knowing the input (voltage or current) to a circuit, we can nd an expression for the output either in the complex frequency domain or in the time domain As an example, suppose Vi (s) and Vo (s) are the input and output voltages to a circuit, respectively, in complex frequency notation Then H (s) = Vo (s) Vi (s) (649)
from which we can obtain the output in the complex frequency domain by computing Vo (s) = H (s)Vi (s) (650)
If Vi (s) is a known damped sinusoid, we can then proceed to determine vo (t) by means of the method illustrated earlier in this section
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617 Given the transfer function H (s) = 3(s + 2)/(s 2 + 2s + 3) and the input Vi (s) = 4 0 , nd the forced response vo (t) if a s = 1 b s = 1 + j 1 c s = 2 + j 1 618 Given the transfer function H (s) = 2(s + 4)/(s 2 + 4s + 5) and the input Vi (s) = 6 30 , nd the forced response vo (t) if a s = 4 + j 1 b s = 2 + j 2
The Laplace Transform The Laplace transform, named after the French mathematician and astronomer Pierre Simon de Laplace, is de ned by L[f (t)] = F (s) =
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